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  1. /*3の倍数*/
  2. //nagaoka daigo
  3. #include <stdio.h>
  4. int main(void) {
  5. 		int s,n,x;
  6. 		s=0;n=0;x=0;
  7. 		do{
  8. 				n+=1;
  9. 				if(n%3==0){
  10. 					x+=1;
  11. 					s+=n;
  12. 				}
  13. 		}while(n<100);
  14. 		printf("3の倍数の個数=%d\n",x);
  15. 		printf("3の倍数の合計=%d\n",s);
  16. 	return 0;
  17. }
  18.  
Success #stdin #stdout 0s 4520KB
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Standard input is empty
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3の倍数の個数=33
3の倍数の合計=1683
https://ideone.com/wGFKAO
language:
C (gcc 8.3)
created:
4 years ago
visibility:
public

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